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\begin{document}

\title{高等代数一}
\subtitle{10-初等矩阵与初等变换}
%\institute{上海立信会计金融学院}
\author{{\ppr LQW}}
%\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
\date{{\ppr 2022年10月25日} }

\maketitle

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\begin{enumerate}
\item  初等矩阵的定义
\item  二阶和三阶的初等矩阵
\item  初等矩阵与初等变换的关系
\item  相抵标准形
\item  寻找初等变换
\item  寻找初等矩阵
\end{enumerate}

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\begin{frame}{10.1. 初等矩阵的定义、二阶初等矩阵 }

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\begin{itemize}
\item  {\color{red}定义：对单位矩阵进行一次初等变换，得到的矩阵称为初等矩阵。}

\item  例子1：写出所有二阶的初等矩阵。

\item  解答：对二阶单位矩阵分别进行三类行初等变换，可得
{\footnotesize 
\begin{eqnarray*}
E_2 = \begin{pmatrix}1&0 \\ 0&1 \end{pmatrix}
&\xrightarrow[\text{ }]{\text{交换第1行和第2行 }}&
\begin{pmatrix}0&1 \\ 1&0 \end{pmatrix} = P_{12}.  \\ 
%
E_2 = \begin{pmatrix}1&0 \\ 0&1 \end{pmatrix}
&\xrightarrow[\text{ }]{\text{第1行乘以非零常数$c$ }}&
\begin{pmatrix}c&0 \\ 0&1 \end{pmatrix} = D_{1}(c). \\ 
%
E_2 = \begin{pmatrix}1&0 \\ 0&1 \end{pmatrix}
&\xrightarrow[\text{ }]{\text{第2行乘以非零常数$c$ }}&
\begin{pmatrix}1&0 \\ 0&c \end{pmatrix} = D_{2}(c). 
\end{eqnarray*}
}

\end{itemize}

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\begin{itemize}

\item  例子1的解答（续）：
{\footnotesize 
\begin{eqnarray*}
E_2 = \begin{pmatrix}1&0 \\ 0&1 \end{pmatrix}
&\xrightarrow[\text{ }]{\text{第1行乘以常数$k$加到第2行 }}&
\begin{pmatrix}1&0 \\ k&1 \end{pmatrix} = T_{21}(k). \\ 
E_2 = \begin{pmatrix}1&0 \\ 0&1 \end{pmatrix}
&\xrightarrow[\text{ }]{\text{第2行乘以常数$k$加到第1行 }}&
\begin{pmatrix}1&k \\ 0&1 \end{pmatrix} = T_{12}(k). 
\end{eqnarray*}
}
\item  因此二阶初等矩阵具有下述形式：
{\footnotesize 
\begin{eqnarray*}
 \begin{pmatrix}0&1 \\ 1&0 \end{pmatrix}, \hspace{0.2cm} 
\begin{pmatrix}c&0 \\ 0&1 \end{pmatrix},  \hspace{0.2cm} 
\begin{pmatrix}1&0 \\ 0&c \end{pmatrix}, \hspace{0.2cm} 
\begin{pmatrix}1&0 \\ k&1 \end{pmatrix}, \hspace{0.2cm} 
\begin{pmatrix}1&k \\ 0&1 \end{pmatrix}. 
\end{eqnarray*}
}
其中 $c\neq 0$. {\color{red}这些矩阵分别称为第一、二、三类初等矩阵。}
\end{itemize}

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\begin{itemize}

\item  例子2：写出所有三阶初等矩阵。

\item  解答：第一、二、三类初等矩阵分别如下，其中 $c\neq 0$, 
{\footnotesize 
\begin{eqnarray*}
&\begin{pmatrix}0&1&0 \\ 1&0&0 \\ 0&0&1 \end{pmatrix},  \hspace{0.2cm} 
\begin{pmatrix}0&0&1 \\ 0&1&0 \\ 1&0&0 \end{pmatrix},  \hspace{0.2cm} 
\begin{pmatrix}1&0&0 \\ 0&0&1 \\ 0&1&0 \end{pmatrix}; &\\ 
&\begin{pmatrix}c&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix},  \hspace{0.2cm} 
\begin{pmatrix}1&0&0 \\ 0&c&0 \\ 0&0&1 \end{pmatrix},  \hspace{0.2cm} 
\begin{pmatrix}1&0&0 \\ 0&1&0 \\ 0&0&c \end{pmatrix}; &\\  
&\begin{pmatrix}1&0&0 \\ k&1&0 \\ 0&0&1 \end{pmatrix},  \hspace{0.2cm} 
\begin{pmatrix}1&0&0 \\ 0&1&0 \\ k&0&1 \end{pmatrix},  \hspace{0.2cm} 
\begin{pmatrix}1&0&0 \\ 0&1&0 \\ 0&k&1 \end{pmatrix},  \hspace{0.2cm} 
\begin{pmatrix}1&k&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix},  \hspace{0.2cm} 
\begin{pmatrix}1&0&k \\ 0&1&0 \\ 0&0&1 \end{pmatrix},  \hspace{0.2cm} 
\begin{pmatrix}1&0&0 \\ 0&1&k \\ 0&0&1 \end{pmatrix}. &
\end{eqnarray*}
}


\end{itemize}

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{\footnotesize 
\begin{eqnarray*}
E_3 = \begin{pmatrix}1&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}
&\xrightarrow[\text{ }]{\text{交换第1行和第2行 }}&
\begin{pmatrix}0&1&0 \\ 1&0&0 \\ 0&0&1 \end{pmatrix} =: P_{12}. \\ 
%
E_3 = \begin{pmatrix}1&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}
&\xrightarrow[\text{ }]{\text{第1行乘以$c$ ($c\neq 0$) }}&
\begin{pmatrix}c&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix} =: D_{1}(c). \\ 
%
E_3 = \begin{pmatrix}1&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}
&\xrightarrow[\text{ }]{\text{第1行乘以$k$加到第2行 }}&
\begin{pmatrix}1&0&0 \\ k&1&0 \\ 0&0&1 \end{pmatrix} =: T_{21}(k).
\end{eqnarray*}
}

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{\small 
\begin{table}[ht]
\centering
%\caption{French cardinal and ordinal numbers 1-10}
%\vspace{0.15cm}
\begin{tabular}{|c|c|c|}\hline
numbers & cardinal& ordinal \\ \hline
1&un, une&premier, première\\ \hline
2&deux&deuxième\\ \hline
3&trois&troisième\\ \hline
4&quatre&quatrième\\ \hline
5&cinq&cinquième\\ \hline
6&six&sixième\\ \hline
7&sept&septième\\ \hline
8&huit&huitième\\ \hline
9&neuf&neuvième\\ \hline
10&dix&dixième\\ \hline
\end{tabular}
\end{table}
}

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\begin{itemize}

\item  {\color{red}定理：
(1)对矩阵 $A$ 进行一次行初等变换，相当于在矩阵 $A$ 的左边乘以一个相应的初等矩阵。
(2)对矩阵 $A$ 进行一次列初等变换，相当于在矩阵 $A$ 的右边乘以一个相应的初等矩阵。
}

\item  证明：对第一类初等变换进行验证，
{\footnotesize 
\begin{eqnarray*}
A &=& 
\begin{pmatrix}a&b&c \\ u&v&w \\ x&y&z \end{pmatrix}
\xrightarrow[\text{ }]{\text{交换第1行和第2行 }}
\begin{pmatrix} u&v&w \\ a&b&c \\ x&y&z \end{pmatrix}, \\
P_{12}\cdot A &=&
\begin{pmatrix}0&1&0 \\ 1&0&0 \\ 0&0&1 \end{pmatrix}
\begin{pmatrix}a&b&c \\ u&v&w \\ x&y&z \end{pmatrix}
=\begin{pmatrix} u&v&w \\ a&b&c \\ x&y&z \end{pmatrix}.
\end{eqnarray*}
}
再验证其余的初等变换，发现对矩阵 $A$ 作{\color{red}行（列）初等变换}与在矩阵 $A$ 的{\color{red}左边（右边）乘以初等矩阵}的效果总是一样的。


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\begin{itemize}

\item  例子3：设矩阵如下，计算矩阵乘积 $PA$, 并找出 $A$ 与 $PA$ 的关系。
{\footnotesize 
\begin{eqnarray*}
A = \begin{pmatrix}a&b&c \\ u&v&w \\ x&y&z \end{pmatrix}, \hspace{0.2cm} 
P = \begin{pmatrix}1&0&0 \\ 0&1&k \\ 0&0&1 \end{pmatrix}. 
\end{eqnarray*}
}

\item  解答：
{\footnotesize 
\begin{eqnarray*}
PA = 
\begin{pmatrix}1&0&0 \\ 0&1&k \\ 0&0&1 \end{pmatrix}
\begin{pmatrix}a&b&c \\ u&v&w \\ x&y&z \end{pmatrix}
= \begin{pmatrix}a&b&c \\ u+kx&v+ky&w+kz \\ x&y&z \end{pmatrix}. 
\end{eqnarray*}
}
将矩阵 $A$ 的第三行乘以 $k$ 加到第二行，就得到矩阵 $PA$. 

\end{itemize}

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\begin{itemize}

\item  例子4：设矩阵如下，计算矩阵乘积 $AP$, 并找出 $A$ 与 $AP$ 的关系。
{\footnotesize 
\begin{eqnarray*}
A = \begin{pmatrix}a&b&c \\ u&v&w \\ x&y&z \end{pmatrix}, \hspace{0.2cm} 
P = \begin{pmatrix}1&0&0 \\ 0&1&k \\ 0&0&1 \end{pmatrix}. 
\end{eqnarray*}
}

\item  解答：
{\footnotesize 
\begin{eqnarray*}
AP = 
\begin{pmatrix}a&b&c \\ u&v&w \\ x&y&z \end{pmatrix}
\begin{pmatrix}1&0&0 \\ 0&1&k \\ 0&0&1 \end{pmatrix}
= 
\begin{pmatrix}a&b&c+bk \\ u&v&w+vk \\ x&y&z+yk \end{pmatrix}. 
\end{eqnarray*}
}
将矩阵 $A$ 的第二列乘以 $k$ 加到第三列，就得到矩阵 $AP$. 

\end{itemize}

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\begin{itemize}

\item  {\color{red}定理：设 $A$ 是一个 $m\times n$ 阶的矩阵，记 $r=R(A)$ 是矩阵 $A$ 的秩。则}
\begin{enumerate}
\item  {\color{red}存在一系列的行初等变换和列初等变换，使得 
{\footnotesize 
\begin{eqnarray*}
A \xrightarrow[\text{ }]{\text{一系列的初等变换 }}
\begin{pmatrix} E_r&O_{r,n-r} \\ O_{m-r,r}&O_{m-r,n-r} \end{pmatrix}. 
\end{eqnarray*}
}}
\item  {\color{red}存在 $m$ 阶的初等矩阵 $P_1, \cdots, P_s$ 与 $n$ 阶的初等矩阵 $Q_1, \cdots, Q_t$ 使得 
{\footnotesize 
\begin{eqnarray*}
P_s \cdots P_1A Q_1\cdots  Q_t 
=
\begin{pmatrix} E_r&O_{r,n-r} \\ O_{m-r,r}&O_{m-r,n-r} \end{pmatrix}. 
\end{eqnarray*}
}}
\end{enumerate}

\item  证明：
%\begin{enumerate}
%\item  
先通过一些行初等变换将矩阵 $A$ 化为行最简形，再通过一些列初等变换得到定理中的形式。
%\item  
根据初等变换与初等矩阵的关系，可知存在定理中的初等矩阵。
%\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  例子5：找到一些初等变换实现下述变换，
{\footnotesize 
\begin{eqnarray*}
A=\begin{pmatrix}1&2&3&4 \\ 5&6&7&8 \\ 9&10&11&12 \end{pmatrix}
\xrightarrow[\text{ }]{\text{一系列的初等变换 }}
\begin{pmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&0&0 \end{pmatrix}=B. 
\end{eqnarray*}
}

\item  解答思路：先用{\color{red}行初等变换}将矩阵 $A$ 化为{\color{red}行最简形}。
{\footnotesize 
\begin{eqnarray*}
A=\begin{pmatrix}1&2&3&4 \\ 5&6&7&8 \\ 9&10&11&12 \end{pmatrix}
\xrightarrow[\text{ }]{\text{行初等变换 }}
\begin{pmatrix} 1&0&-1&-2 \\ 0&1&2&3 \\ 0&0&0&0 \end{pmatrix}
\xrightarrow[\text{ }]{\text{列初等变换 }}
\begin{pmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&0&0 \end{pmatrix}=B. 
\end{eqnarray*}
}

\end{itemize}

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\begin{itemize}

\item  例子6：在例子5中，找到一些初等矩阵 $P_1, \cdots, P_s$ 与 $Q_1, \cdots, Q_t$ 使得 
{\footnotesize 
\begin{eqnarray*}
P_s \cdots P_1A Q_1\cdots  Q_t =B. 
\end{eqnarray*}
}
\item  解答思路：将每一步{\color{red}初等变换}对应的{\color{red}初等矩阵}写出来。例如，
{\footnotesize 
\begin{eqnarray*}
\begin{pmatrix}1&0&0 \\ 0&1&0 \\ -9&0&1 \end{pmatrix}
\begin{pmatrix}1&0&0 \\ -5&1&0 \\ 0&0&1 \end{pmatrix}
\begin{pmatrix}1&2&3&4 \\ 5&6&7&8 \\ 9&10&11&12 \end{pmatrix}
=
\begin{pmatrix} 1&2&3&4 \\ 0&-4&-8&-12 \\ 0&-8&-16&-24 \end{pmatrix}. 
\end{eqnarray*}
}

\vfill 

\item  注：因为有不同的初等变换的途径达到相同的目标，所以这些初等矩阵不是唯一确定的。

\end{itemize}

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\begin{itemize}

\item  例子7：通过第三类行和列的初等变换，实现矩阵的下述初等变换，
{\footnotesize 
\begin{eqnarray*}
A=\begin{pmatrix}3&2&1 \\ 4&9&8 \\ 5&6&7 \end{pmatrix}
\xrightarrow[\text{ }]{\text{一系列的第三类初等变换 }}
\begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&d \end{pmatrix}=B. 
\end{eqnarray*}
}

\item  解答思路：为避免分式运算，我们先将第二列乘以 $(-1)$ 加到第一列，然后再用 $(1,1)$ 位置上的 $1$ 消去其下方的数字，得到
{\footnotesize 
\begin{eqnarray*}
A=\begin{pmatrix}3&2&1 \\ 4&9&8 \\ 5&6&7 \end{pmatrix}
\xrightarrow[\text{ }]{\text{一个列初等变换 }}
\begin{pmatrix} 1&2&1 \\ -5&9&8 \\ -1&6&7 \end{pmatrix}
\xrightarrow[\text{ }]{\text{两个行初等变换 }}
\begin{pmatrix} 1&2&1 \\ 0&19&13 \\ 0&8&8 \end{pmatrix}. 
\end{eqnarray*}
}


\end{itemize}

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\begin{enumerate}

\item  设有下述三个初等矩阵，
{\footnotesize 
\begin{eqnarray*}
A=\begin{pmatrix}1&0&0 \\ a&1&0 \\ 0&0&1 \end{pmatrix},  \hspace{0.2cm} 
B=\begin{pmatrix}1&0&0 \\ 0&1&0 \\ b&0&1 \end{pmatrix},  \hspace{0.2cm} 
C=\begin{pmatrix}1&0&0 \\ 0&1&0 \\ 0&c&1 \end{pmatrix}. 
\end{eqnarray*}
}
分别计算矩阵的乘积 $AB, ABC, BA, CBA$. 你能找到更快的计算方法吗？


\end{enumerate}

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\begin{enumerate}

\item  注意到初等矩阵与初等变换的关系，可以简化矩阵的乘法运算。
{\footnotesize 
\begin{eqnarray*}
&AB=\begin{pmatrix}1&0&0 \\ a&1&0 \\ b&0&1 \end{pmatrix},  \hspace{0.2cm} 
ABC=\begin{pmatrix}1&0&0 \\ a&1&0 \\ b&c&1 \end{pmatrix},  & \\ 
&BA=\begin{pmatrix}1&0&0 \\ a&1&0 \\ b&0&1 \end{pmatrix},  \hspace{0.2cm} 
CBA=\begin{pmatrix}1&0&0 \\ a&1&0 \\ b+ca&c&1 \end{pmatrix}.  &
\end{eqnarray*}
}

\end{enumerate}

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